Find the explicit formula for the nth term of the sequence?

[Topgun_68]

Homework Statement

Consider the sequence n₁, n₂, n₃, ... that satisfies the recurrence relation n_k = n_k-1 / k + 1 for all integers k ≥ 2 with the initial condition that n₁ = 1. Find the explicit formula n_k for the nth term of the sequence?


2. The attempt at a solution

I calculated out the terms for n = 1, n = 2, n = 3 & n = 4 but there is no obvious relation that I can see because of the decimal numbers. For example:

Starting variable n₁ = 1 so..

n₂ = \frac{1}{3} = .333

n₃ = \frac{.333}{4} = .08325

n₄ = \frac{.08325}{5} = .01665

Any hints on how to calculate the explicit formula for n_k?
Thanks for any info!

 

[SteamKing]

What do you notice about the relationship between the denominators and the index k?
What relationship do you deduce about the numerators and n-sub-k?

 

[Topgun_68]

Hmm, the denominators are increasing sequentially by 1 more than k.

As the index goes up the numerator is getting smaller.

I'll keep playing with the number until something jumps out at me. I don't know where my instructor comes up with these tough problems, but I can do the easy ones in the book

What do you notice about the relationship between the denominators and the index k?
What relationship do you deduce about the numerators and n-sub-k?

 

[HallsofIvy]

Homework Statement

Consider the sequence n₁, n₂, n₃, ... that satisfies the recurrence relation n_k = n_k-1 / k + 1 for all integers k ≥ 2 with the initial condition that n₁ = 1. Find the explicit formula n_k for the nth term of the sequence?2. The attempt at a solution

I calculated out the terms for n = 1, n = 2, n = 3 & n = 4 but there is no obvious relation that I can see because of the decimal numbers. For example:

Starting variable n₁ = 1 so..

n₂ = \frac{1}{3} = .333

It would have been better NOT to approximate with a decimal

n₃ = \frac{.333}{4} = .08325

n₃= (1/3)/(3+1)= 1/(3(4))/

n₄ = \frac{.08325}{5} = .01665

n₄= (1/(3(4)))/(4+ 1)= 11/(3(4)(5))

Any hints on how to calculate the explicit formula for n_k?
Thanks for any info!

1/3, 1/(3(4)), 1/(3(4)(5)), should remind you of a factorial.

 

[Topgun_68]

Ah, I never would have noticed with the decimal numbers. I won't make that mistake again. It looks like the factorial starts at 3 so is he equation correct below? I get correct answers with a numerator of 2 but can't figure out how to get the 2? I just guessed it. Thanks for the help!


n_k = \frac{2}{(n+1)n!}

It would have been better NOT to approximate with a decimal


n₃= (1/3)/(3+1)= 1/(3(4))/


n₄= (1/(3(4)))/(4+ 1)= 11/(3(4)(5))


1/3, 1/(3(4)), 1/(3(4)(5)), should remind you of a factorial.

 

[Ray Vickson]

Homework Statement

Consider the sequence n₁, n₂, n₃, ... that satisfies the recurrence relation n_k = n_k-1 / k + 1 for all integers k ≥ 2 with the initial condition that n₁ = 1. Find the explicit formula n_k for the nth term of the sequence?


2. The attempt at a solution

I calculated out the terms for n = 1, n = 2, n = 3 & n = 4 but there is no obvious relation that I can see because of the decimal numbers. For example:

Starting variable n₁ = 1 so..

n₂ = \frac{1}{3} = .333

n₃ = \frac{.333}{4} = .08325

n₄ = \frac{.08325}{5} = .01665

Any hints on how to calculate the explicit formula for n_k?
Thanks for any info!

What you wrote means
n_k = \frac{n_{k-1}}{k} + 1.
Did you mean that, or did you mean
n_k = \frac{n_{k-1}}{k+1} ?
If you mean the latter, USE PARENTHESES, like this: n_k = n_k-1/(k+1).

Anyway, do not use limited decimal representations; 1/3 is NOT 0.333, 1/12 is NOT 0.8325 (it is actually .08333333333333333 ...) The errors you get are building up more and more as you increase the number of steps, and using decimals like that serves no useful purpose here.

 

[Mentallic]

Ah, I never would have noticed with the decimal numbers. I won't make that mistake again. It looks like the factorial starts at 3 so is he equation correct below? I get correct answers with a numerator of 2 but can't figure out how to get the 2? I just guessed it. Thanks for the help!n_k = \frac{2}{(n+1)n!}

Since you have

n_k = \frac{1}{3.4.5...k}

and

\frac{1}{k!}=\frac{1}{2.3.4.5...k}=\frac{1}{2}\cdot\frac{1}{3.4.5...k}=\frac{1}{2}\cdot n_k

Does this manipulation make sense so far?

So finally, what is n_k?EDIT: Also, you guessed the formula to be

\frac{2}{(n+1)n!}

which I should note is equivalent to

=\frac{2}{(n+1)!}

And also, you're looking for the kth term, not the nth term. Your formula should be in terms of k.

n_k = f(k)

So if you're looking for n₂ then you plug k=2 into the formula.

 

[Topgun_68]

Sorry, I am trying to learn how to put the equations in using the correct symbols, so I forgot the parenthesis, which I know totally changes the equation. I meant the later..

n_k = \frac{(n_{k-1})}{(k+1)} ?

What you wrote means
n_k = \frac{n_{k-1}}{k} + 1.
Did you mean that, or did you mean
n_k = \frac{n_{k-1}}{k+1} ?
If you mean the latter, USE PARENTHESES, like this: n_k = n_k-1/(k+1).

Anyway, do not use limited decimal representations; 1/3 is NOT 0.333, 1/12 is NOT 0.8325 (it is actually .08333333333333333 ...) The errors you get are building up more and more as you increase the number of steps, and using decimals like that serves no useful purpose here.

 

[Topgun_68]

Thanks for all the clarifications. I have been mixing up the terms and it's been hurting my grades.

so would it be

n_k = \frac{2}{(k+1)!}

Thanks. That's what I like about this forums. You can always count on everyone to correct all these errors. I have been learning a lot from them

Since you have

n_k = \frac{1}{3.4.5...k}

and

\frac{1}{k!}=\frac{1}{2.3.4.5...k}=\frac{1}{2}\cdot\frac{1}{3.4.5...k}=\frac{1}{2}\cdot n_k

Does this manipulation make sense so far?

So finally, what is n_k?


EDIT: Also, you guessed the formula to be

\frac{2}{(n+1)n!}

which I should note is equivalent to

=\frac{2}{(n+1)!}

And also, you're looking for the kth term, not the nth term. Your formula should be in terms of k.

n_k = f(k)

So if you're looking for n₂ then you plug k=2 into the formula.

 

[Mentallic]

so would it be

n_k = \frac{2}{(k+1)!}

No. Look back to my derivation. I ended up with

\frac{1}{k!}=\frac{1}{2}\cdot n_k

so then

n_k = \frac{2}{k!}

Did it make sense to you?

 

[Topgun_68]

Yes I can see how you did all the math to acquire the answer, but how you got the \frac{1}{2} confuses me a little. Is it because the question stated k ≥ 2 so you pulled it out of the factorial? Than you multiply both sides by 2 to isolate the n_k.

Thanks for your patience with me!

No. Look back to my derivation. I ended up with

\frac{1}{k!}=\frac{1}{2}\cdot n_k

so then

n_k = \frac{2}{k!}

Did it make sense to you?

 

[Mentallic]

Sorry for misleading you, but I just wrote down the pattern for myself and noticed I did make a big blunder, you were right, it's

n_k = \frac{2}{(k+1)!}

My derivation in the earlier post still follows the same procedure, except that we begin with

n_k = \frac{1}{3.4.5...k(k+1)}

So then

\frac{1}{(k+1)!} =\frac{1}{2.3.4.5...k(k+1)} =\frac{1}{2} \frac{1}{3.4.5...k(k+1)} = \frac{1}{2}n_k

Hence we have that

n_k = \frac{2}{(k+1)!}

Sorry about that.

Yes I can see how you did all the math to acquire the answer, but how you got the \frac{1}{2} confuses me a little. Is it because the question stated k ≥ 2 so you pulled it out of the factorial? Than you multiply both sides by 2 to isolate the n_k.

Thanks for your patience with me!

The factor of 1/2 came from the fact that when we noticed the pattern for n_k to be

n_k = \frac{1}{3.4.5...k(k+1)}

The denominator is nearly

(k+1)! = 2.3.4.5...k(k+1)

But we're missing the 2, so we put in there! But if we multiplied the denominator by 2, then in order to keep it equal, we need to multiply the numerator by 2 as well:

\frac{1}{3.4.5...(k+1)}=\frac{2}{2}\cdot \frac{1}{3.4.5...(k+1)} = \frac{2}{2.3.4...(k+1)}=\frac{2}{(k+1)!}

If the question was instead changed to "for integers k\geq 1" then our initial condition would have to be on n₀, so let n₀=1, then

n_1=\frac{n_0}{1+1}=\frac{1}{2}

n_2=\frac{n_1}{2+1}=\frac{1/2}{3}=\frac{1}{2.3}

Following this pattern, we can see that kth term will be

n_k = \frac{1}{2.3.4...k(k+1)}

Which is simply

n_k = \frac{1}{(k+1)!}

 

[Topgun_68]

Ah, that's where the two came from. It cancels out the almost factorial

Thanks again for everyone's assistance. I like how everyone on here helps you figure out the answer for yourself, which is turn makes it's easier to understand come exam time.

Sorry for misleading you, but I just wrote down the pattern for myself and noticed I did make a big blunder, you were right, it's

n_k = \frac{2}{(k+1)!}

My derivation in the earlier post still follows the same procedure, except that we begin with

n_k = \frac{1}{3.4.5...k(k+1)}

So then

\frac{1}{(k+1)!} =\frac{1}{2.3.4.5...k(k+1)} =\frac{1}{2} \frac{1}{3.4.5...k(k+1)} = \frac{1}{2}n_k

Hence we have that

n_k = \frac{2}{(k+1)!}

Sorry about that.



The factor of 1/2 came from the fact that when we noticed the pattern for n_k to be

n_k = \frac{1}{3.4.5...k(k+1)}

The denominator is nearly

(k+1)! = 2.3.4.5...k(k+1)

But we're missing the 2, so we put in there! But if we multiplied the denominator by 2, then in order to keep it equal, we need to multiply the numerator by 2 as well:

\frac{1}{3.4.5...(k+1)}=\frac{2}{2}\cdot \frac{1}{3.4.5...(k+1)} = \frac{2}{2.3.4...(k+1)}=\frac{2}{(k+1)!}

If the question was instead changed to "for integers k\geq 1" then our initial condition would have to be on n₀, so let n₀=1, then

n_1=\frac{n_0}{1+1}=\frac{1}{2}

n_2=\frac{n_1}{2+1}=\frac{1/2}{3}=\frac{1}{2.3}

Following this pattern, we can see that kth term will be

n_k = \frac{1}{2.3.4...k(k+1)}

Which is simply

n_k = \frac{1}{(k+1)!}